The pitch of a screw gauge is 1 mm and there are 100 "divisions" on circular scale. When faces A and B are just touching each without putting anything between the studs 32nd divisions of the circular scale (below its Zero) coincides with the reference line. When a glass plate is placed between the studs, the linear scale reads 4 divisions and the circular reads 16 divisions. Find the thickness of the glass plate. Zero of linnear scale is not hidden from circular scale when A and B touches each other.
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mujee nhi pta
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Given :
Linear scale reading = 4mm
Pitch of screw guage = 1mm
n = 32
To find :
The thickness of the glass plate
Solution :
- Least count (L.C) = Pitch / Number of circular scale divisions
=1/100 mm
L.C = 0.01 mm
- Error = n×L.C =32×0.01
- e=0.32mm
- Circular scale reading = 16×0.01 =0.16mm
- Measured readind = 4+0.16 =4.16mm
- Total reading = 4.16 - e
=4.16 - 0.32
=3.84mm
The thickness of the glass slab is 3.84mm
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